Aahh!No, there is a voltage drop in the diode that needs to be overcome, then the diode starts conducting. The above graph using the Red curve is roughly 1.5V. The 1.5V has to be supplied before the current starts to flow, after that the additional voltage across the diode's internal resistance determines the current.
Short recap:
- V_f needs to be supplied so I can put current through the diode, that voltage does not produce light emission.
- to put a desired current through then, I take the internal resistance, calculated as the slope from the linear part of the diode-specific I_f/V_f graph, R_diode=V_f/I_f (R_diode is a fix material constant)
- the voltage needed for that/related to that current is calculated by rearranging the equation to V=R_diode*I, the overall V needed on that line is the sum of forward voltage to overcome diodes internal resistance V_f and V related to desired current
- wattage is I²*R (square of throughput current times internal resistance)
- since voltage is
very helpful, thanks a lot!3.1 V, 3.0 V, 3.05 V, 2.9 V, As I showed with the graph, there is a diode drop and a resistance drop in a Led (actually any diode). Just to make things simple let us assume the internal resistance of each of these Leds are the same and it is just the diode drop that is different. And let's say that once the diode drop voltage is supplied it only takes 0.1 V to get 20 mA through the Led. So a quick approximation, say the first diode needs 3.0 V to cover the diode voltage drop. That means 0.1V / 20 mA = 5 ohm. The math for the others is close enough, we will call all of them 5 ohms.
And say we turn up the power supply until it supplies 3.1 V. That means the first diode is passing 20 mA. The second Led has a diode drop of (3.0 - 0.1 V = 2.9 V) so it has 3.1 V - 2.9 V = 0.2 V across its 5 ohm resistance. 0.2 V / 5 ohms = 40 mA.
The third one, 3.1 V - 2.95 V = 0.15 V. 0.15 V / 5 ohms = 30 mA
The fourth, 3.1 V - 2.8 V = 0.3 V. 0.3 V / 5 ohms = 60 mA.
The power supply would be supplying 20 + 40 +30 + 60 = 150 mA.
So it is not good to put parallel Leds across a constant voltage power supply without balancing resistors inline with them. These resistors are much bigger than the internal resistance and it balances out the currents. It does come at a cost of added voltage and power wasted in them. But that is life.
Now what happens if you have a constant current power supply and you have it set for 80 mA? I will not do the math, just to say the 2.9 V and 3.0 V Led will get most of the current and the 3.05 V 3.1V Leds will get a fraction of the current and will barely turn on if at all.
The different currents in a parallel wired CV circuit as calculated above (20 + 40 +30 + 60 = 150 mA) would be no problem, if the LEDs' optima would lie exactly or in tolerable range of these currents?
and http://lednique.com is a good source too!
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